Weld strength (BS EN 1993-1-8)
Fillet weld check, directional method (BS EN 1993-1-8 4.5.3.2)
- where
- normal stress on the throat, equal to τ⊥ (N/mm²)
- design force per mm of weld (N/mm)
- angle between force and weld axis (°)
- throat thickness (mm)
- shear stress along the weld (N/mm²)
- equivalent stress (N/mm²)
- utilisation, combined check
- ultimate tensile strength of the weaker part (N/mm²)
- correlation factor
- partial factor
- utilisation, normal stress check
Explain this equation
Checks a fillet weld by the directional method in BS EN 1993-1-8, which gives credit for welds loaded across their length. Use it when the simplified method is too conservative.
The force per millimetre is split into a part across the weld (sin φ) and a part along it (cos φ). For an equal-leg fillet the transverse part divides equally into normal and shear stress on the 45° throat, each F sin φ / (a√2). The longitudinal part is pure shear along the throat. These combine into an equivalent stress, with shear counting three times, which is compared with the weld strength. A separate check limits the normal stress alone.
This is the von Mises style stress combination used in Eurocode 3 for weld throats.
Sense check: 1000 N/mm transverse on an a5 fillet gives about 141 N/mm² normal stress, an equivalent stress of about 283 and a utilisation of about 0.68 for S355. The same force along the weld gives 0.83.
Watch out: this assumes an equal-leg fillet with the force in the plane at angle φ. Combine forces from different directions before entering them.
Checks an equal-leg fillet weld in a 90° joint under a force at any angle to the weld, by resolving it into stresses on the throat plane.
Enter your values to see the result.
Good to know
For a 90° equal-leg fillet the throat plane is at 45°, so a transverse force splits equally into a normal stress σ⊥ and a shear τ⊥, each F/(a√2); a longitudinal force gives shear τ∥ = F/a. Both checks must be at or below 1.0. A purely transverse load gives about 1.225 times the simplified-method resistance, matching the K factor in SCI guidance.
Common mistake
Applying the 1.225 transverse bonus when the force direction is uncertain or reverses. If in doubt, use the simplified method.
Limits
Equal-leg fillet in a 90° joint, force per unit length uniform along the weld and acting in the plane perpendicular to the attached plate. Moment and eccentric connections need the force per mm resolved first. Design by a competent structural engineer.
Sources
- BS EN 1993-1-8:2005 Eurocode 3: Design of steel structures – Part 1-8: Design of joints (4.5.3.2 directional method, 4.5.3.3 simplified method, Table 4.1 correlation factor βw) and UK National Annex (γM2 = 1.25)
- ArcelorMittal/SCI Orange Book: Bolts and welds (S355 fu = 470 N/mm², βw = 0.9, γM2 = 1.25, Fw,L,Rd = fvw,d × a, transverse Fw,T,Rd = K × Fw,L,Rd with K = 1.225)
- SCI Advisory Desk AD 399: design of partial penetration butt welds (S355 fu = 470 N/mm², βw = 0.9, γM2 = 1.25; directional method checks)
Results are calculations from the published methods and figures shown, and depend on the data you enter. Follow the welding procedure specification (WPS), BS EN ISO 15614/9606, BS EN 1090, BS EN 1993 and manufacturers’ data. Structural welds need qualified welders and engineered designs.